Another week, another problem.
Problem 88: Connected components
We know that a DFS will find the entire connected component containing the beginning node before finishing. Therefore we can use the solution from problem 87 straightforwardly:
connected_components :: (Eq a, Show a) => Graph a -> [[a]] connected_components (Graph vs es) = foldr f  vs where -- Flatten the accumulator in order to determine if `v` belongs to a connected -- component which we already enumerated: f v acc = if v `elem` (concat acc) then acc else dfs (Graph vs es) v : acc
Problem 89: Bipartite graphs
A graph is bipartite if and only if it is 2-colorable. This fairly intuitive hypothesis has been proven by Asratian, Armen, Denley, Tristan and Häggkvist in “Bipartite Graphs and their Applications”, from Cambridge Tracts in Mathematics 131.
Since we have already implemented graph colorisation, we will use it thusly:
bipartite :: (Eq a, Ord a, Show a) => Graph a -> Bool bipartite g = 2 == (maximum $ map snd $ wpcolor g)
Easy, right? :-)
Problem 90: Eight queens problem
The assumption to use a list of integers, where each element of a list represents a
column, makes this very easy to implement. By filtering all permutations of the range
[1..n], we can solve it in a straight-forward, but slow, manner:
import Data.List (permutations) queens :: Int -> [[Int]] -- All permutations, i.e. queens' placements must be "safe": -- This test is slow-ish. To do better, I would write a list comprehension that generates -- permutations and incorporates the `safe` test as a guard, to fail as early as possible: queens n = filter test (permutations [1..n]) where test  = True test (q : qs) = safe q qs && test qs -- A placement of queens is "safe" iff there are no two queens in the same row or -- in the same diagonal: safe p ps = not (p `elem` ps || same_diagonal p ps) -- Two pieces (r1, c1) and (r2, c2) are on the same diagonal iff abs(r2 - r1) == -- abs (c2 - c1), i.e. the distance between their columns is equal to the distance -- between their rows. -- Columns are given the same numbering as the rows, i.e. [1..n] (n is omitted for -- no reason): same_diagonal p ps = any (\(distance, row) -> abs (row - p) == distance) $ zip [1..] ps main = print $ queens 8
Notice that, by construction, we will have exactly one queen per column and no more, so we don’t need to incorporate that extra test in our code. Neat!
Problem 91: Knight’s tour
Another instance where a wiki solution is wrong. That one never terminates, as far as I could tell.
import Data.Ord (comparing) import Data.List ((\\), minimumBy) type Square = (Int, Int) -- Is a square part of a NxN board? on_board :: Int -> Square -> Bool on_board n (x, y) = 1 <= x && x <= n && 1 <= y && y <= n -- Valid moves for a knight for a given square, on a NxN board: valid_moves :: Int -> Square -> [Square] valid_moves n (x, y) = filter (on_board n) [(x+1, y+2), (x+1, y-2), (x+2, y+1), (x+2, y-1), (x-1, y+2), (x-1, y-2), (x-2, y+1), (x-2, y-1)] knights_to :: Int -> Square -> [Square] knights_to n finish = knights' [finish] where -- Our search ends when there's no more moves: knights' moves | next_choices ==  = moves -- Otherwise we keep going by picking a `next` move: | otherwise = knights' (next : moves) where -- The choice for our next move is a heuristic by Warnsdorf, which says -- to prefer squares with fewer onward moves. We won't be concerned with -- how to break possible ties: next = minimumBy (comparing (length . choices)) next_choices -- We pick valid moves from the last move taken, which is the head of -- `moves`: next_choices = choices (head moves) -- All the valid moves from a given square, to which we haven't been to -- yet: choices square = valid_moves n square \\ moves
Short pause here, maybe to solve the last few problems over the weekend. Cheers, folks!« Past Future »